Consider the grid of $\Bbb R^3$ that passes through the points of $\Bbb Z^3$:
$$\{(x,y,z)\in\Bbb R^3:\text{ at least two of }x,y,z\text{ are integers}\}.$$
My question is if it is possible to embed the unknot into this grid so that the three product projections $\Bbb R^3\rightarrow \Bbb R^2$ are graphs with no cycles, i.e., trees.
I've only found embeddings where only two of the three projections are trees, so this might be false, but I still don't know.
Any hint or ideas are really appreciated.
This puzzle appeared in Mathematical Mind-Benders, by Peter Winkler, a collection of the interesting mathematical puzzles. Your puzzle (without the grid restriction) was in the chapter "Severe Challenges," and most of the puzzles were hard enough already! There is an excellent discussion of the history of the problem in that book, and also in the introduction of paper: https://arxiv.org/pdf/1507.02355.pdf.
Anyways, here is the simplest solution, which appears on the cover of the Winkler book.
Even more interesting, this blog gives a solution where the embedding is a trefoil knot: